a.

Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)

\(=\dfrac{2}{15}\)

b.

\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)

Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)

\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)

c.

\(\int\limits^2_1\dfrac{dx}{\sqrt{4-x^2}}\)

Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=\dfrac{\pi}{6}\\x=2\Rightarrow u=\dfrac{\pi}{2}\end{matrix}\right.\)

\(I=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{\sqrt{4-4sin^2u}}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{2cosu}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}du\)

\(=u|^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}=\dfrac{\pi}{3}\)

Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)

\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)

Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)

\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)

I=A+B=...

Câu a)

\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)

Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)

Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)

Vậy :

\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)

\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)

Câu b)

\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)

\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)

Do đó:

\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)

Câu c)

\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)

\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)

\(=\frac{x^2}{2}+c+\ln ^2x\)

\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)

Câu d)

\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)

\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)

\(=1+\ln 3\)

Câu e)

Xét \(\int 3x(x+\sqrt{x^2+16})dx=\int 3x^2dx+\int 3x\sqrt{x^2+16}dx\)

Có:

\(\int 3x^2dx=x^3+c\)

\(\int 3x\sqrt{x^2+16}dx=\frac{3}{2}\int \sqrt{x^2+16}d(x^2+16)\)

\(=\sqrt{(x^2+16)^3}+c\)

Do đó: \(\int 3x(x+\sqrt{x^2+16})dx=x^3+\sqrt{(x^2+16)^3}+c\)

\(\Rightarrow \int ^{3}_{0}3x(x+\sqrt{x^2+16})dx=\left.\begin{matrix} 3\\ 0\end{matrix}\right|(x^3+\sqrt{(x^2+16)^3}+c)=88\)

\(t=\sqrt{x+1}>0\Rightarrow x=t^2-1\)

\(\Rightarrow dt=\dfrac{1}{2\sqrt{x+1}}dx=\dfrac{1}{2t}dx\Rightarrow dx=2tdt\)

\(\Rightarrow I=\int\dfrac{2t^4-t^2-1}{t}.2tdt=2\int\left(2t^4-t^2-1\right)dt\)

Đến đây bạn làm bình thường r thay t bằng căn(x+1) vô là được

a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)

b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)

Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)

\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)

\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)

c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)

d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)

e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)

\(\dfrac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{\left(x+1\right)^2x-x^2\left(x+1\right)}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{x\left(x+1\right)}\)

\(=\dfrac{\sqrt{x}}{x}-\dfrac{\sqrt{x+1}}{x+1}=x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\)

Do đó:

\(I=\int\limits^2_1\left[x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\right]dx=\left(2\sqrt{x}-2\sqrt{x+1}\right)|^2_1=...\)

\(I=\int\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}=\int\sqrt{x+1}dx+\int\sqrt{x}dx\)

Xet \(I_1=\int\sqrt{x+1}dx\)

\(t=x+1\Rightarrow dt=dx\Rightarrow I_1=\int\sqrt{t}.dt=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+C\)

\(\Rightarrow I=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}x^{\dfrac{3}{2}}+C\)

P/s: Bạn tự thay cận vô ạ